# find the principal unit normal vector to the curve

• Jun 14th 2012, 04:27 PM
icelated
find the principal unit normal vector to the curve
find the principal unit normal vector to the curve at the specific value of the parameter.

$\displaystyle r(t) = ti + \frac {1}{2}t^2j, t = 2$

i need to use this formula

$\displaystyle T(t) = \frac {r'(t)}{ || r'(t) ||}$

so, to find r'(t)

$\displaystyle r'(t) = i + t j$

so, now to find the magnitude: simplified

$\displaystyle || r'(t) || = \sqrt {1 + t^2}$

puting them together

$\displaystyle T(t) = \frac {i + t j} {\sqrt {1 + t^2}}$

From here is where i get lost

I need to find T'(t)

I can seem to follow the solutions manual. I think they are using the quiotient rule due to the denominator having the 3/2
fraction. But if they are i still cant see how they are doing it.
Can anyone help me work out the next step for T'(t)?

$\displaystyle T'(t) = \frac { -t}{(t^2 + 1)^{3/2}}i + \frac {1}{ (t^2 + 1) ^{3/2}} j$

Thank you so much!!!
• Jun 14th 2012, 07:37 PM
Reckoner
Re: find the principal unit normal vector to the curve
$\displaystyle \mathbf{T}(t) = \frac{\mathbf{i} + t\mathbf{j}}{\sqrt{1 + t^2}}$

$\displaystyle \Rightarrow\mathbf{T}(t) = \left(1+t^2\right)^{-1/2}\mathbf{i} + t\left(1+t^2\right)^{-1/2}\mathbf{j}$

Differentiating,

$\displaystyle \mathbf{T}'(t) = -\frac12(2t)\left(1+t^2\right)^{-3/2}\mathbf{i} + \left[-\frac12t(2t)\left(1+t^2\right)^{-3/2} + \left(1+t^2\right)^{-1/2}\right]\mathbf{j}$

$\displaystyle \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \left[\frac{-t^2}{\left(1+t^2\right)^{3/2}} + \frac1{\left(1+t^2\right)^{1/2}}\right]\mathbf{j}$

$\displaystyle \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac{-t^2+\left(1+t^2\right)}{\left(1+t^2\right)^{3/2}}\mathbf{j}$

$\displaystyle \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}$

We find the magnitude,

$\displaystyle \left\|\mathbf{T}'(t)\right\| = \sqrt{\frac{t^2+1}{\left(1+t^2\right)^3}$

$\displaystyle \Rightarrow\left\|\mathbf{T}'(t)\right\| = \sqrt{\frac1{\left(t^2+1\right)^2}$

$\displaystyle \Rightarrow\left\|\mathbf{T}'(t)\right\| = \frac1{t^2+1}$

And therefore,

$\displaystyle \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\left\|\mathbf{T}'(t)\right \|}$

$\displaystyle =\left(t^2+1\right)\left[\frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}\right]$

$\displaystyle =\frac{-t}{\sqrt{1+t^2}}\mathbf{i} + \frac1{\sqrt{1+t^2}}\mathbf{j}\right]$
• Jun 14th 2012, 07:44 PM
icelated
Re: find the principal unit normal vector to the curve
in your second step the first arrow before Differentiating, im not sure how you got j

for i did you just bring up the denominator? doesnt make sense
• Jun 14th 2012, 07:47 PM
Reckoner
Re: find the principal unit normal vector to the curve
Quote:

Originally Posted by icelated
in your second step the first arrow before Differentiating, im not sure how you got j

for i did you just bring up the denominator? doesnt make sense

Did you forget these properties of exponents from algebra?

$\displaystyle x^{-n} = \frac1{x^n}$

$\displaystyle x^{m/n} = \sqrt[n]{x^m}$
• Jun 15th 2012, 07:19 AM
icelated
Re: find the principal unit normal vector to the curve
Wow, okay how did i miss that! I wasnt splitting the i and j up. Once i did now i can see the answer fall out.